Question

Consider the following reactions giving major product. Identify the correct reaction. 

• A. Reaction 1
• B. Reaction 2
• C. Reaction 3
• D. Reaction 4

Hint:

Hofmann bromamide reaction is identified by loss of one carbon atom when an amide is converted into a primary amine.

Answer 1

The Correct Option is D

Concept: To identify the correct reaction giving the stated major product, we must analyze:
The nature of the reactants
The reaction conditions
The well-known named reactions involved
Whether the product formation is chemically consistent
Step 1: Analyze Reaction (1) Reaction (1) shows potassium phthalimide reacting with an aryl halide, followed by heat and hydrolysis, to give aniline. This resembles the {Gabriel phthalimide synthesis}. However:
Gabriel synthesis works efficiently for {alkyl halides}, not aryl halides.
Aryl halides do not undergo \(S_N2\) substitution easily. \[ \Rightarrow Reaction (1) is incorrect. \]
Step 2: Analyze Reaction (2) This reaction involves benzylamine treated with \(\mathrm{CHCl_3/KOH}\), followed by reduction.
\(\mathrm{CHCl_3/KOH}\) causes the {carbylamine reaction}, which requires a {primary amine}.
The product is an isocyanide, which does not revert back to the original amine upon reduction in this manner. \[ \Rightarrow Reaction (2) is incorrect. \]
Step 3: Analyze Reaction (3) This reaction shows acetanilide undergoing nitration using a mixture of \(\mathrm{HNO_3/H_2SO_4}\).
The \(-NHCO-\) group is an ortho–para directing group.
Nitration gives a mixture of ortho- and para-nitroacetanilide, with the para isomer as the major product.
The reaction as drawn does not correctly represent the major product distribution. \[ \Rightarrow Reaction (3) is incorrect. \]
Step 4: Analyze Reaction (4) Reaction (4) shows: \[ \mathrm{CH_3CH_2CONH_2 + Br_2 + 4KOH \xrightarrow{\Delta} CH_3CH_2NH_2 + 2KBr + K_2CO_3 + 2H_2O} \] This is a classic example of the {Hofmann bromamide reaction}. Key features of Hofmann bromamide reaction:
Converts an amide into a primary amine
The amine formed has {one carbon less} than the original amide
Strong base and bromine are required Here:
Propionamide (\(\mathrm{CH_3CH_2CONH_2}\)) gives ethylamine (\(\mathrm{CH_3CH_2NH_2}\))
This matches the expected product exactly \[ \Rightarrow Reaction (4) is correct. \] Final Answer: \[ \boxed{\text{Reaction (4)}} \]